The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((94, 97)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-94)^2+(y_2-97)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(90,94)
2
(38,55)
3
(99,109)
4
(64,57)
5
(114,76)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-94)^2+(y_2-97)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(90-94)^2+(94-97)^2}~=~5\]
\[\sqrt{(38-94)^2+(55-97)^2}~=~70\]
\[\sqrt{(99-94)^2+(109-97)^2}~=~13\]
\[\sqrt{(64-94)^2+(57-97)^2}~=~50\]
\[\sqrt{(114-94)^2+(76-97)^2}~=~29\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((102, 105)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-102)^2+(y_2-105)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(126,112)
2
(142,114)
3
(22,45)
4
(120,81)
5
(105,109)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-102)^2+(y_2-105)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(126-102)^2+(112-105)^2}~=~25\]
\[\sqrt{(142-102)^2+(114-105)^2}~=~41\]
\[\sqrt{(22-102)^2+(45-105)^2}~=~100\]
\[\sqrt{(120-102)^2+(81-105)^2}~=~30\]
\[\sqrt{(105-102)^2+(109-105)^2}~=~5\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((103, 98)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-98)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(43,123)
2
(43,178)
3
(113,122)
4
(52,30)
5
(58,74)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-98)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(43-103)^2+(123-98)^2}~=~65\]
\[\sqrt{(43-103)^2+(178-98)^2}~=~100\]
\[\sqrt{(113-103)^2+(122-98)^2}~=~26\]
\[\sqrt{(52-103)^2+(30-98)^2}~=~85\]
\[\sqrt{(58-103)^2+(74-98)^2}~=~51\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((91, 101)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-91)^2+(y_2-101)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(61,173)
2
(131,92)
3
(154,161)
4
(103,66)
5
(109,181)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-91)^2+(y_2-101)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(61-91)^2+(173-101)^2}~=~78\]
\[\sqrt{(131-91)^2+(92-101)^2}~=~41\]
\[\sqrt{(154-91)^2+(161-101)^2}~=~87\]
\[\sqrt{(103-91)^2+(66-101)^2}~=~37\]
\[\sqrt{(109-91)^2+(181-101)^2}~=~82\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((104, 107)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-104)^2+(y_2-107)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(92,116)
2
(83,35)
3
(71,163)
4
(124,59)
5
(8,135)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-104)^2+(y_2-107)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(92-104)^2+(116-107)^2}~=~15\]
\[\sqrt{(83-104)^2+(35-107)^2}~=~75\]
\[\sqrt{(71-104)^2+(163-107)^2}~=~65\]
\[\sqrt{(124-104)^2+(59-107)^2}~=~52\]
\[\sqrt{(8-104)^2+(135-107)^2}~=~100\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((92, 109)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-92)^2+(y_2-109)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(35,33)
2
(71,37)
3
(48,142)
4
(107,101)
5
(148,151)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-92)^2+(y_2-109)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(35-92)^2+(33-109)^2}~=~95\]
\[\sqrt{(71-92)^2+(37-109)^2}~=~75\]
\[\sqrt{(48-92)^2+(142-109)^2}~=~55\]
\[\sqrt{(107-92)^2+(101-109)^2}~=~17\]
\[\sqrt{(148-92)^2+(151-109)^2}~=~70\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((107, 95)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-107)^2+(y_2-95)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(83,25)
2
(75,71)
3
(95,100)
4
(131,105)
5
(11,67)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-107)^2+(y_2-95)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(83-107)^2+(25-95)^2}~=~74\]
\[\sqrt{(75-107)^2+(71-95)^2}~=~40\]
\[\sqrt{(95-107)^2+(100-95)^2}~=~13\]
\[\sqrt{(131-107)^2+(105-95)^2}~=~26\]
\[\sqrt{(11-107)^2+(67-95)^2}~=~100\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((99, 109)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-99)^2+(y_2-109)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(144,169)
2
(87,93)
3
(115,46)
4
(95,112)
5
(129,125)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-99)^2+(y_2-109)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(144-99)^2+(169-109)^2}~=~75\]
\[\sqrt{(87-99)^2+(93-109)^2}~=~20\]
\[\sqrt{(115-99)^2+(46-109)^2}~=~65\]
\[\sqrt{(95-99)^2+(112-109)^2}~=~5\]
\[\sqrt{(129-99)^2+(125-109)^2}~=~34\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((95, 102)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-95)^2+(y_2-102)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(35,57)
2
(149,30)
3
(86,62)
4
(172,138)
5
(115,123)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-95)^2+(y_2-102)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(35-95)^2+(57-102)^2}~=~75\]
\[\sqrt{(149-95)^2+(30-102)^2}~=~90\]
\[\sqrt{(86-95)^2+(62-102)^2}~=~41\]
\[\sqrt{(172-95)^2+(138-102)^2}~=~85\]
\[\sqrt{(115-95)^2+(123-102)^2}~=~29\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((105, 95)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-105)^2+(y_2-95)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(90,87)
2
(54,27)
3
(181,152)
4
(121,125)
5
(98,71)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-105)^2+(y_2-95)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(90-105)^2+(87-95)^2}~=~17\]
\[\sqrt{(54-105)^2+(27-95)^2}~=~85\]
\[\sqrt{(181-105)^2+(152-95)^2}~=~95\]
\[\sqrt{(121-105)^2+(125-95)^2}~=~34\]
\[\sqrt{(98-105)^2+(71-95)^2}~=~25\]
Question
Using Desmos, determine which of the given points are within a distance of 11 units from point (3, 5).
Graph the inequality \((x-3)^2+(y-5)^2~\le~11^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(12,11)
(11,14)
(7,16)
(-1,14)
(-5,12)
(-8,2)
(-7,1)
(-2,-5)
(8,-5)
(15,2)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (12,11) is in
point (11,14) is NOT in
point (7,16) is NOT in
point (-1,14) is in
point (-5,12) is in
point (-8,2) is NOT in
point (-7,1) is in
point (-2,-5) is NOT in
point (8,-5) is NOT in
point (15,2) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 11 units from point (6, 4).
Graph the inequality \((x-6)^2+(y-4)^2~\le~11^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(17,9)
(9,13)
(4,15)
(-1,11)
(-3,10)
(-3,0)
(-3,-5)
(5,-7)
(12,-6)
(18,2)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (17,9) is NOT in
point (9,13) is in
point (4,15) is NOT in
point (-1,11) is in
point (-3,10) is in
point (-3,0) is in
point (-3,-5) is NOT in
point (5,-7) is NOT in
point (12,-6) is NOT in
point (18,2) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 8 units from point (-6, 1).
Graph the inequality \((x+6)^2+(y-1)^2~\le~8^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(3,3)
(-2,7)
(-6,8)
(-9,7)
(-12,4)
(-14,-2)
(-9,-6)
(-3,-8)
(2,-2)
(2,-1)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (3,3) is NOT in
point (-2,7) is in
point (-6,8) is in
point (-9,7) is in
point (-12,4) is in
point (-14,-2) is NOT in
point (-9,-6) is in
point (-3,-8) is NOT in
point (2,-2) is NOT in
point (2,-1) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 15 units from point (5, -1).
Graph the inequality \((x-5)^2+(y+1)^2~\le~15^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(18,9)
(16,9)
(4,14)
(-6,10)
(-10,-2)
(-4,-13)
(3,-17)
(9,-14)
(20,-7)
(20,-3)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (18,9) is NOT in
point (16,9) is in
point (4,14) is NOT in
point (-6,10) is NOT in
point (-10,-2) is NOT in
point (-4,-13) is in
point (3,-17) is NOT in
point (9,-14) is in
point (20,-7) is NOT in
point (20,-3) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 8 units from point (-2, -3).
Graph the inequality \((x+2)^2+(y+3)^2~\le~8^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(6,-3)
(4,5)
(1,6)
(-6,4)
(-10,2)
(-11,-4)
(-9,-5)
(-6,-9)
(2,-10)
(5,-8)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (6,-3) is in
point (4,5) is NOT in
point (1,6) is NOT in
point (-6,4) is NOT in
point (-10,2) is NOT in
point (-11,-4) is NOT in
point (-9,-5) is in
point (-6,-9) is in
point (2,-10) is NOT in
point (5,-8) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 15 units from point (6, -8).
Graph the inequality \((x-6)^2+(y+8)^2~\le~15^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(18,-1)
(15,5)
(3,5)
(-8,1)
(-9,-8)
(-9,-9)
(-2,-22)
(9,-22)
(21,-14)
(20,-9)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (18,-1) is in
point (15,5) is NOT in
point (3,5) is in
point (-8,1) is NOT in
point (-9,-8) is in
point (-9,-9) is NOT in
point (-2,-22) is NOT in
point (9,-22) is in
point (21,-14) is NOT in
point (20,-9) is in
Question
Using Desmos, determine which of the given points are within a distance of 10 units from point (3, 2).
Graph the inequality \((x-3)^2+(y-2)^2~\le~10^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(12,6)
(2,12)
(-2,11)
(-6,8)
(-6,1)
(-2,-6)
(0,-7)
(5,-8)
(4,-7)
(11,-3)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (12,6) is in
point (2,12) is NOT in
point (-2,11) is NOT in
point (-6,8) is NOT in
point (-6,1) is in
point (-2,-6) is in
point (0,-7) is in
point (5,-8) is NOT in
point (4,-7) is in
point (11,-3) is in
Question
Using Desmos, determine which of the given points are within a distance of 15 units from point (7, 4).
Graph the inequality \((x-7)^2+(y-4)^2~\le~15^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(21,7)
(17,15)
(2,18)
(-8,8)
(-7,-4)
(-1,-8)
(1,-8)
(4,-9)
(13,-11)
(20,-4)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (21,7) is in
point (17,15) is in
point (2,18) is in
point (-8,8) is NOT in
point (-7,-4) is NOT in
point (-1,-8) is in
point (1,-8) is in
point (4,-9) is in
point (13,-11) is NOT in
point (20,-4) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 15 units from point (8, -4).
Graph the inequality \((x-8)^2+(y+4)^2~\le~15^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(24,-3)
(16,9)
(2,10)
(-4,4)
(-7,-1)
(-4,-12)
(8,-19)
(9,-18)
(17,-14)
(22,-5)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (24,-3) is NOT in
point (16,9) is NOT in
point (2,10) is NOT in
point (-4,4) is in
point (-7,-1) is NOT in
point (-4,-12) is in
point (8,-19) is in
point (9,-18) is in
point (17,-14) is in
point (22,-5) is in
Question
Using Desmos, determine which of the given points are within a distance of 13 units from point (-2, -7).
Graph the inequality \((x+2)^2+(y+7)^2~\le~13^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(12,-7)
(11,-5)
(2,6)
(-8,4)
(-13,-2)
(-15,-3)
(-15,-8)
(-10,-17)
(0,-21)
(6,-15)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
As the first item, use the standard form of a circle to visualize the first criterion.
\[(x-30)^2+(y-62)^2~\le~28^2\]
As the second item, use the standard form of a circle to visualize the second criterion.
\[(x-53)^2+(y-84)^2~\le~12^2\]
As the third item, copy/paste the list of points. Click the “Label” checkbox.
Read which points are in the overlapping region.
point (45,76) is in
point (48,70) is NOT in
point (49,87) is NOT in
point (46,80) is in
point (44,85) is in
point (40,79) is NOT in
point (55,75) is NOT in
point (41,71) is NOT in
point (42,69) is NOT in
point (56,81) is NOT in
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (103, 92).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (91, 95).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (108, 105).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (100, 101).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (99, 107).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (101, 103).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (101, 91).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (104, 93).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (106, 98).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (108, 106).
Question
Consider the triangle with vertices at the following coordinates:
(6, 18)
(9, 7)
(15, 3)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{3^2+11^2}~=~ 11.4017543\]
\[\sqrt{6^2+4^2}~=~ 7.2111026\]
\[\sqrt{9^2+15^2}~=~ 17.4928557\]
To find the perimeter, add up the three lengths.
\[P~=~11.4017543+7.2111026+17.4928557\]
\[P~=~36.1057125\]
Question
Consider the triangle with vertices at the following coordinates:
(17, 16)
(4, 2)
(14, 7)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{13^2+14^2}~=~ 19.1049732\]
\[\sqrt{10^2+5^2}~=~ 11.1803399\]
\[\sqrt{3^2+9^2}~=~ 9.486833\]
To find the perimeter, add up the three lengths.
\[P~=~19.1049732+11.1803399+9.486833\]
\[P~=~39.772146\]
Question
Consider the triangle with vertices at the following coordinates:
(18, 9)
(5, 19)
(3, 12)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{13^2+10^2}~=~ 16.4012195\]
\[\sqrt{2^2+7^2}~=~ 7.2801099\]
\[\sqrt{15^2+3^2}~=~ 15.2970585\]
To find the perimeter, add up the three lengths.
\[P~=~16.4012195+7.2801099+15.2970585\]
\[P~=~38.9783879\]
Question
Consider the triangle with vertices at the following coordinates:
(19, 16)
(11, 9)
(17, 4)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{8^2+7^2}~=~ 10.6301458\]
\[\sqrt{6^2+5^2}~=~ 7.8102497\]
\[\sqrt{2^2+12^2}~=~ 12.1655251\]
To find the perimeter, add up the three lengths.
\[P~=~10.6301458+7.8102497+12.1655251\]
\[P~=~30.6059205\]
Question
Consider the triangle with vertices at the following coordinates:
(16, 1)
(6, 14)
(11, 17)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{10^2+13^2}~=~ 16.4012195\]
\[\sqrt{5^2+3^2}~=~ 5.8309519\]
\[\sqrt{5^2+16^2}~=~ 16.7630546\]
To find the perimeter, add up the three lengths.
\[P~=~16.4012195+5.8309519+16.7630546\]
\[P~=~38.995226\]
Question
Consider the triangle with vertices at the following coordinates:
(2, 5)
(4, 13)
(8, 3)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{2^2+8^2}~=~ 8.2462113\]
\[\sqrt{4^2+10^2}~=~ 10.7703296\]
\[\sqrt{6^2+2^2}~=~ 6.3245553\]
To find the perimeter, add up the three lengths.
\[P~=~8.2462113+10.7703296+6.3245553\]
\[P~=~25.3410962\]
Question
Consider the triangle with vertices at the following coordinates:
(13, 5)
(6, 10)
(15, 14)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{7^2+5^2}~=~ 8.6023253\]
\[\sqrt{9^2+4^2}~=~ 9.8488578\]
\[\sqrt{2^2+9^2}~=~ 9.2195445\]
To find the perimeter, add up the three lengths.
\[P~=~8.6023253+9.8488578+9.2195445\]
\[P~=~27.6707275\]
Question
Consider the triangle with vertices at the following coordinates:
(15, 5)
(8, 1)
(2, 7)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{7^2+4^2}~=~ 8.0622577\]
\[\sqrt{6^2+6^2}~=~ 8.4852814\]
\[\sqrt{13^2+2^2}~=~ 13.1529464\]
To find the perimeter, add up the three lengths.
\[P~=~8.0622577+8.4852814+13.1529464\]
\[P~=~29.7004856\]
Question
Consider the triangle with vertices at the following coordinates:
(17, 9)
(6, 7)
(10, 18)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{11^2+2^2}~=~ 11.1803399\]
\[\sqrt{4^2+11^2}~=~ 11.7046999\]
\[\sqrt{7^2+9^2}~=~ 11.4017543\]
To find the perimeter, add up the three lengths.
\[P~=~11.1803399+11.7046999+11.4017543\]
\[P~=~34.286794\]
Question
Consider the triangle with vertices at the following coordinates:
(9, 18)
(15, 8)
(2, 13)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taken differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{6^2+10^2}~=~ 11.6619038\]
\[\sqrt{13^2+5^2}~=~ 13.9283883\]
\[\sqrt{7^2+5^2}~=~ 8.6023253\]
To find the perimeter, add up the three lengths.
\[P~=~11.6619038+13.9283883+8.6023253\]
\[P~=~34.1926173\]
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
-36
63
-33
41
65
18
-5
33
-20
-58
-3
73
-47
-17
54
-21
91
37
-19
-9
-46
10
47
-21
94
2
95
2
-16
-40
-59
-36
-22
-27
-45
-10
-22
4
-44
-76
-8
-20
75
-29
-86
50
16
-42
47
69
53
-68
39
-25
-74
49
-49
10
-15
51
29
9
21
50
-22
1
-31
-62
14
-17
47
-30
How many of these arrows landed in the ring worth 2 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 2 points has an inner radius of 80 mm and an outer radius of 90 mm. So, we want to know how many arrows have distances (from origin) between 80 mm and 90 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 2 points.
I’ll assume you have pasted the first arrow’s position in the second row, so -36 in cell A2 and 63 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 80 mm away, use =IF(C2>80,1,0) in cell D2.
To check if the arrow is less than 90 mm away, use =IF(C2<90,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 80 mm
is less than 90 mm
is in ring
how many in
-36
63
72.56032
0
1
0
5
-33
41
52.63079
0
1
0
65
18
67.44627
0
1
0
-5
33
33.37664
0
1
0
-20
-58
61.35145
0
1
0
-3
73
73.06162
0
1
0
-47
-17
49.98000
0
1
0
54
-21
57.93962
0
1
0
91
37
98.23441
1
0
0
-19
-9
21.02380
0
1
0
-46
10
47.07441
0
1
0
47
-21
51.47815
0
1
0
94
2
94.02127
1
0
0
95
2
95.02105
1
0
0
-16
-40
43.08132
0
1
0
-59
-36
69.11584
0
1
0
-22
-27
34.82815
0
1
0
-45
-10
46.09772
0
1
0
-22
4
22.36068
0
1
0
-44
-76
87.81799
1
1
1
-8
-20
21.54066
0
1
0
75
-29
80.41144
1
1
1
-86
50
99.47864
1
0
0
16
-42
44.94441
0
1
0
47
69
83.48653
1
1
1
53
-68
86.21485
1
1
1
39
-25
46.32494
0
1
0
-74
49
88.75246
1
1
1
-49
10
50.01000
0
1
0
-15
51
53.16014
0
1
0
29
9
30.36445
0
1
0
21
50
54.23099
0
1
0
-22
1
22.02272
0
1
0
-31
-62
69.31811
0
1
0
14
-17
22.02272
0
1
0
47
-30
55.75841
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 2 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
79
21
51
-41
-45
21
-47
18
-9
-54
-7
-49
20
27
48
-30
-21
69
38
60
25
13
92
-18
-26
38
-13
9
8
38
46
17
-21
-39
-33
4
-21
17
-28
23
76
38
4
-3
-10
-9
-33
42
-21
29
-3
-32
-33
-27
-34
-3
21
30
-41
-21
-2
-53
63
-11
81
-14
-35
27
22
3
-4
4
How many of these arrows landed in the ring worth 2 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 2 points has an inner radius of 80 mm and an outer radius of 90 mm. So, we want to know how many arrows have distances (from origin) between 80 mm and 90 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 2 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 79 in cell A2 and 21 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 80 mm away, use =IF(C2>80,1,0) in cell D2.
To check if the arrow is less than 90 mm away, use =IF(C2<90,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 80 mm
is less than 90 mm
is in ring
how many in
79
21
81.743501
1
1
1
3
51
-41
65.436993
0
1
0
-45
21
49.658836
0
1
0
-47
18
50.328918
0
1
0
-9
-54
54.744863
0
1
0
-7
-49
49.497475
0
1
0
20
27
33.600595
0
1
0
48
-30
56.603887
0
1
0
-21
69
72.124892
0
1
0
38
60
71.021124
0
1
0
25
13
28.178006
0
1
0
92
-18
93.744333
1
0
0
-26
38
46.043458
0
1
0
-13
9
15.811388
0
1
0
8
38
38.832976
0
1
0
46
17
49.040799
0
1
0
-21
-39
44.294469
0
1
0
-33
4
33.241540
0
1
0
-21
17
27.018512
0
1
0
-28
23
36.235342
0
1
0
76
38
84.970583
1
1
1
4
-3
5.000000
0
1
0
-10
-9
13.453624
0
1
0
-33
42
53.413481
0
1
0
-21
29
35.805028
0
1
0
-3
-32
32.140317
0
1
0
-33
-27
42.638011
0
1
0
-34
-3
34.132096
0
1
0
21
30
36.619667
0
1
0
-41
-21
46.065171
0
1
0
-2
-53
53.037722
0
1
0
63
-11
63.953108
0
1
0
81
-14
82.200973
1
1
1
-35
27
44.204072
0
1
0
22
3
22.203603
0
1
0
-4
4
5.656854
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 2 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
-9
21
-43
-2
26
30
52
0
23
15
-29
-16
-21
-29
-32
12
18
8
3
-50
29
62
-52
-33
-57
-16
-34
70
36
-24
-25
-19
-64
-51
-29
18
42
-60
-28
-26
-16
-8
44
-24
-39
8
33
21
-20
-34
-8
-69
5
3
-15
-6
2
-5
-67
43
-3
-23
8
-9
-6
23
-48
44
-50
11
-22
31
How many of these arrows landed in the ring worth 9 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 9 points has an inner radius of 10 mm and an outer radius of 20 mm. So, we want to know how many arrows have distances (from origin) between 10 mm and 20 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 9 points.
I’ll assume you have pasted the first arrow’s position in the second row, so -9 in cell A2 and 21 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 10 mm away, use =IF(C2>10,1,0) in cell D2.
To check if the arrow is less than 20 mm away, use =IF(C2<20,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 10 mm
is less than 20 mm
is in ring
how many in
-9
21
22.847319
1
0
0
4
-43
-2
43.046487
1
0
0
26
30
39.698867
1
0
0
52
0
52.000000
1
0
0
23
15
27.459060
1
0
0
-29
-16
33.120990
1
0
0
-21
-29
35.805028
1
0
0
-32
12
34.176015
1
0
0
18
8
19.697716
1
1
1
3
-50
50.089919
1
0
0
29
62
68.447060
1
0
0
-52
-33
61.587336
1
0
0
-57
-16
59.203041
1
0
0
-34
70
77.820306
1
0
0
36
-24
43.266615
1
0
0
-25
-19
31.400637
1
0
0
-64
-51
81.835200
1
0
0
-29
18
34.132096
1
0
0
42
-60
73.239334
1
0
0
-28
-26
38.209946
1
0
0
-16
-8
17.888544
1
1
1
44
-24
50.119856
1
0
0
-39
8
39.812058
1
0
0
33
21
39.115214
1
0
0
-20
-34
39.446166
1
0
0
-8
-69
69.462220
1
0
0
5
3
5.830952
0
1
0
-15
-6
16.155494
1
1
1
2
-5
5.385165
0
1
0
-67
43
79.611557
1
0
0
-3
-23
23.194827
1
0
0
8
-9
12.041595
1
1
1
-6
23
23.769729
1
0
0
-48
44
65.115282
1
0
0
-50
11
51.195703
1
0
0
-22
31
38.013156
1
0
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 9 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
21
-6
-28
40
-40
-55
9
3
-31
-7
14
21
-27
-14
-3
52
34
-47
10
-32
47
-4
-54
-69
-12
24
18
63
0
6
-11
63
12
53
-15
57
-47
-4
5
-19
62
-65
-38
-12
47
49
-26
14
-22
-97
60
10
14
-40
18
-47
9
-56
-43
51
-8
0
20
-46
-9
35
-68
31
38
-4
20
7
How many of these arrows landed in the ring worth 5 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 5 points has an inner radius of 50 mm and an outer radius of 60 mm. So, we want to know how many arrows have distances (from origin) between 50 mm and 60 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 5 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 21 in cell A2 and -6 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 50 mm away, use =IF(C2>50,1,0) in cell D2.
To check if the arrow is less than 60 mm away, use =IF(C2<60,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 50 mm
is less than 60 mm
is in ring
how many in
21
-6
21.840330
0
1
0
7
-28
40
48.826223
0
1
0
-40
-55
68.007352
1
0
0
9
3
9.486833
0
1
0
-31
-7
31.780497
0
1
0
14
21
25.238859
0
1
0
-27
-14
30.413813
0
1
0
-3
52
52.086467
1
1
1
34
-47
58.008620
1
1
1
10
-32
33.526109
0
1
0
47
-4
47.169906
0
1
0
-54
-69
87.618491
1
0
0
-12
24
26.832816
0
1
0
18
63
65.520989
1
0
0
0
6
6.000000
0
1
0
-11
63
63.953108
1
0
0
12
53
54.341513
1
1
1
-15
57
58.940648
1
1
1
-47
-4
47.169906
0
1
0
5
-19
19.646883
0
1
0
62
-65
89.827613
1
0
0
-38
-12
39.849718
0
1
0
47
49
67.896981
1
0
0
-26
14
29.529646
0
1
0
-22
-97
99.463561
1
0
0
60
10
60.827625
1
0
0
14
-40
42.379240
0
1
0
18
-47
50.328918
1
1
1
9
-56
56.718604
1
1
1
-43
51
66.708320
1
0
0
-8
0
8.000000
0
1
0
20
-46
50.159745
1
1
1
-9
35
36.138622
0
1
0
-68
31
74.732858
1
0
0
38
-4
38.209946
0
1
0
20
7
21.189620
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 5 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
1
40
21
33
-1
-74
31
-14
-27
-51
18
76
-57
-60
32
9
-29
-54
8
14
-70
54
-35
-42
20
-66
1
-10
11
-19
-6
7
32
-26
17
-65
5
-26
32
40
-33
-23
58
39
-37
-16
-8
11
-38
19
79
11
15
62
10
31
-49
50
-11
-19
17
-21
-43
21
-3
-4
24
-42
22
30
-14
-23
How many of these arrows landed in the ring worth 4 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 4 points has an inner radius of 60 mm and an outer radius of 70 mm. So, we want to know how many arrows have distances (from origin) between 60 mm and 70 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 4 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 1 in cell A2 and 40 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 60 mm away, use =IF(C2>60,1,0) in cell D2.
To check if the arrow is less than 70 mm away, use =IF(C2<70,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 60 mm
is less than 70 mm
is in ring
how many in
1
40
40.012498
0
1
0
5
21
33
39.115214
0
1
0
-1
-74
74.006756
1
0
0
31
-14
34.014703
0
1
0
-27
-51
57.706152
0
1
0
18
76
78.102497
1
0
0
-57
-60
82.758685
1
0
0
32
9
33.241540
0
1
0
-29
-54
61.294372
1
1
1
8
14
16.124516
0
1
0
-70
54
88.408144
1
0
0
-35
-42
54.671748
0
1
0
20
-66
68.963759
1
1
1
1
-10
10.049876
0
1
0
11
-19
21.954498
0
1
0
-6
7
9.219544
0
1
0
32
-26
41.231056
0
1
0
17
-65
67.186308
1
1
1
5
-26
26.476405
0
1
0
32
40
51.224994
0
1
0
-33
-23
40.224371
0
1
0
58
39
69.892775
1
1
1
-37
-16
40.311289
0
1
0
-8
11
13.601470
0
1
0
-38
19
42.485292
0
1
0
79
11
79.762146
1
0
0
15
62
63.788714
1
1
1
10
31
32.572995
0
1
0
-49
50
70.007143
1
0
0
-11
-19
21.954498
0
1
0
17
-21
27.018512
0
1
0
-43
21
47.853944
0
1
0
-3
-4
5.000000
0
1
0
24
-42
48.373547
0
1
0
22
30
37.202151
0
1
0
-14
-23
26.925824
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 4 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
35
69
-12
51
39
-22
54
-6
-55
21
-85
23
41
22
-32
-55
-22
-16
18
17
10
18
11
12
-23
48
19
20
-22
74
-6
-24
-61
-32
-15
-25
19
-17
45
-38
1
24
7
18
9
-32
18
2
-60
-56
59
-18
60
-1
35
26
9
-44
69
11
-27
-38
-25
6
-35
58
10
-55
22
54
-25
-27
How many of these arrows landed in the ring worth 8 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 8 points has an inner radius of 20 mm and an outer radius of 30 mm. So, we want to know how many arrows have distances (from origin) between 20 mm and 30 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 8 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 35 in cell A2 and 69 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 20 mm away, use =IF(C2>20,1,0) in cell D2.
To check if the arrow is less than 30 mm away, use =IF(C2<30,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 20 mm
is less than 30 mm
is in ring
how many in
35
69
77.36924
1
0
0
9
-12
51
52.39275
1
0
0
39
-22
44.77723
1
0
0
54
-6
54.33231
1
0
0
-55
21
58.87274
1
0
0
-85
23
88.05680
1
0
0
41
22
46.52956
1
0
0
-32
-55
63.63175
1
0
0
-22
-16
27.20294
1
1
1
18
17
24.75884
1
1
1
10
18
20.59126
1
1
1
11
12
16.27882
0
1
0
-23
48
53.22593
1
0
0
19
20
27.58623
1
1
1
-22
74
77.20104
1
0
0
-6
-24
24.73863
1
1
1
-61
-32
68.88396
1
0
0
-15
-25
29.15476
1
1
1
19
-17
25.49510
1
1
1
45
-38
58.89822
1
0
0
1
24
24.02082
1
1
1
7
18
19.31321
0
1
0
9
-32
33.24154
1
0
0
18
2
18.11077
0
1
0
-60
-56
82.07314
1
0
0
59
-18
61.68468
1
0
0
60
-1
60.00833
1
0
0
35
26
43.60046
1
0
0
9
-44
44.91102
1
0
0
69
11
69.87131
1
0
0
-27
-38
46.61545
1
0
0
-25
6
25.70992
1
1
1
-35
58
67.74216
1
0
0
10
-55
55.90170
1
0
0
22
54
58.30952
1
0
0
-25
-27
36.79674
1
0
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 8 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
6
7
-51
28
80
-28
-6
-54
48
-7
-18
41
-43
-26
-41
28
-53
-54
43
-18
-33
-7
3
13
-27
-56
19
-42
1
-37
-46
23
-44
-3
4
54
7
-31
17
-16
-7
-11
78
48
-14
-33
-7
22
-41
-48
49
-5
37
17
35
77
33
13
-7
-50
-19
30
36
42
-30
-26
3
-60
-81
0
18
-6
How many of these arrows landed in the ring worth 4 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 4 points has an inner radius of 60 mm and an outer radius of 70 mm. So, we want to know how many arrows have distances (from origin) between 60 mm and 70 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 4 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 6 in cell A2 and 7 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 60 mm away, use =IF(C2>60,1,0) in cell D2.
To check if the arrow is less than 70 mm away, use =IF(C2<70,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 60 mm
is less than 70 mm
is in ring
how many in
6
7
9.219544
0
1
0
3
-51
28
58.180753
0
1
0
80
-28
84.758480
1
0
0
-6
-54
54.332311
0
1
0
48
-7
48.507731
0
1
0
-18
41
44.777226
0
1
0
-43
-26
50.249378
0
1
0
-41
28
49.648766
0
1
0
-53
-54
75.663730
1
0
0
43
-18
46.615448
0
1
0
-33
-7
33.734256
0
1
0
3
13
13.341664
0
1
0
-27
-56
62.169124
1
1
1
19
-42
46.097722
0
1
0
1
-37
37.013511
0
1
0
-46
23
51.429564
0
1
0
-44
-3
44.102154
0
1
0
4
54
54.147945
0
1
0
7
-31
31.780497
0
1
0
17
-16
23.345235
0
1
0
-7
-11
13.038405
0
1
0
78
48
91.586025
1
0
0
-14
-33
35.846897
0
1
0
-7
22
23.086793
0
1
0
-41
-48
63.126856
1
1
1
49
-5
49.254441
0
1
0
37
17
40.718546
0
1
0
35
77
84.581322
1
0
0
33
13
35.468296
0
1
0
-7
-50
50.487622
0
1
0
-19
30
35.510562
0
1
0
36
42
55.317267
0
1
0
-30
-26
39.698867
0
1
0
3
-60
60.074953
1
1
1
-81
0
81.000000
1
0
0
18
-6
18.973666
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 4 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
-17
7
3
-34
-19
-36
-2
63
47
22
-62
-20
53
10
60
7
27
-13
-30
28
29
11
-46
-17
7
21
-10
-4
-31
-48
-1
-45
30
30
16
-66
-30
31
-14
11
-6
-32
-8
-19
-10
-33
-46
81
-13
-41
-8
3
34
-11
36
-32
67
2
9
-35
-35
-4
-55
1
-18
17
-32
-62
-81
25
-10
-14
How many of these arrows landed in the ring worth 8 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 8 points has an inner radius of 20 mm and an outer radius of 30 mm. So, we want to know how many arrows have distances (from origin) between 20 mm and 30 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 8 points.
I’ll assume you have pasted the first arrow’s position in the second row, so -17 in cell A2 and 7 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 20 mm away, use =IF(C2>20,1,0) in cell D2.
To check if the arrow is less than 30 mm away, use =IF(C2<30,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 20 mm
is less than 30 mm
is in ring
how many in
-17
7
18.384776
0
1
0
4
3
-34
34.132096
1
0
0
-19
-36
40.706265
1
0
0
-2
63
63.031738
1
0
0
47
22
51.894123
1
0
0
-62
-20
65.145990
1
0
0
53
10
53.935146
1
0
0
60
7
60.406953
1
0
0
27
-13
29.966648
1
1
1
-30
28
41.036569
1
0
0
29
11
31.016125
1
0
0
-46
-17
49.040799
1
0
0
7
21
22.135944
1
1
1
-10
-4
10.770330
0
1
0
-31
-48
57.140178
1
0
0
-1
-45
45.011110
1
0
0
30
30
42.426407
1
0
0
16
-66
67.911707
1
0
0
-30
31
43.139309
1
0
0
-14
11
17.804494
0
1
0
-6
-32
32.557641
1
0
0
-8
-19
20.615528
1
1
1
-10
-33
34.481879
1
0
0
-46
81
93.150416
1
0
0
-13
-41
43.011626
1
0
0
-8
3
8.544004
0
1
0
34
-11
35.735137
1
0
0
36
-32
48.166378
1
0
0
67
2
67.029844
1
0
0
9
-35
36.138622
1
0
0
-35
-4
35.227830
1
0
0
-55
1
55.009090
1
0
0
-18
17
24.758837
1
1
1
-32
-62
69.771054
1
0
0
-81
25
84.770278
1
0
0
-10
-14
17.204650
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 8 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
-25
40
-30
-35
-46
-53
18
5
25
-1
-11
34
54
19
-21
77
-13
53
-27
51
-9
30
23
-12
52
-16
-7
44
-6
36
39
13
45
7
-5
-2
31
8
47
-7
13
-15
28
13
-7
55
31
5
15
46
2
-14
-11
34
-37
35
24
-7
-33
28
-79
-41
40
17
-37
-15
-21
-21
27
-48
37
-46
How many of these arrows landed in the ring worth 6 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 6 points has an inner radius of 40 mm and an outer radius of 50 mm. So, we want to know how many arrows have distances (from origin) between 40 mm and 50 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 6 points.
I’ll assume you have pasted the first arrow’s position in the second row, so -25 in cell A2 and 40 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 40 mm away, use =IF(C2>40,1,0) in cell D2.
To check if the arrow is less than 50 mm away, use =IF(C2<50,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 40 mm
is less than 50 mm
is in ring
how many in
-25
40
47.169906
1
1
1
9
-30
-35
46.097722
1
1
1
-46
-53
70.178344
1
0
0
18
5
18.681542
0
1
0
25
-1
25.019992
0
1
0
-11
34
35.735137
0
1
0
54
19
57.245087
1
0
0
-21
77
79.812280
1
0
0
-13
53
54.571055
1
0
0
-27
51
57.706152
1
0
0
-9
30
31.320919
0
1
0
23
-12
25.942244
0
1
0
52
-16
54.405882
1
0
0
-7
44
44.553339
1
1
1
-6
36
36.496575
0
1
0
39
13
41.109610
1
1
1
45
7
45.541190
1
1
1
-5
-2
5.385165
0
1
0
31
8
32.015621
0
1
0
47
-7
47.518417
1
1
1
13
-15
19.849433
0
1
0
28
13
30.870698
0
1
0
-7
55
55.443665
1
0
0
31
5
31.400637
0
1
0
15
46
48.383882
1
1
1
2
-14
14.142136
0
1
0
-11
34
35.735137
0
1
0
-37
35
50.931326
1
0
0
24
-7
25.000000
0
1
0
-33
28
43.278170
1
1
1
-79
-41
89.005618
1
0
0
40
17
43.462628
1
1
1
-37
-15
39.924930
0
1
0
-21
-21
29.698485
0
1
0
27
-48
55.072679
1
0
0
37
-46
59.033889
1
0
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 6 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
-18
3
62
3
21
0
52
-21
10
4
-29
61
-1
-64
5
57
61
17
24
3
17
-6
-34
15
39
1
9
-30
-21
34
12
20
35
8
-2
38
25
-7
10
5
-36
-49
-9
-6
26
-9
10
-27
24
-43
36
-60
-50
-35
5
-13
-21
14
18
-31
81
-45
-9
-62
36
-6
1
-19
-17
22
-1
-32
How many of these arrows landed in the ring worth 7 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 7 points has an inner radius of 30 mm and an outer radius of 40 mm. So, we want to know how many arrows have distances (from origin) between 30 mm and 40 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
where \(\Delta x=x_2-x_1\) and \(\Delta y=y_2-y_1\) and \(\Delta z=z_2-z_1\).
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2+6x+y^2-8y~=~-21\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2+6x+y^2-8y~=~-21\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2+6x+9+y^2-8y+16~=~-21+9+16\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x+3)^2+(y-4)^2~=~4\]
Notice \(4\) is a perfect square integer.
\[(x+3)^2+(y-4)^2~=~2^2\]
Thus,
\[h=-3\]\[k=4\]\[r=2\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-4x+y^2+12y~=~-31\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-4x+y^2+12y~=~-31\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-4x+4+y^2+12y+36~=~-31+4+36\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-2)^2+(y+6)^2~=~9\]
Notice \(9\) is a perfect square integer.
\[(x-2)^2+(y+6)^2~=~3^2\]
Thus,
\[h=2\]\[k=-6\]\[r=3\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-18x+y^2-12y~=~-113\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-18x+y^2-12y~=~-113\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-18x+81+y^2-12y+36~=~-113+81+36\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-9)^2+(y-6)^2~=~4\]
Notice \(4\) is a perfect square integer.
\[(x-9)^2+(y-6)^2~=~2^2\]
Thus,
\[h=9\]\[k=6\]\[r=2\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2+4x+y^2-10y~=~52\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2+4x+y^2-10y~=~52\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2+4x+4+y^2-10y+25~=~52+4+25\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x+2)^2+(y-5)^2~=~81\]
Notice \(81\) is a perfect square integer.
\[(x+2)^2+(y-5)^2~=~9^2\]
Thus,
\[h=-2\]\[k=5\]\[r=9\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-8x+y^2-10y~=~-32\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-8x+y^2-10y~=~-32\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-8x+16+y^2-10y+25~=~-32+16+25\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-4)^2+(y-5)^2~=~9\]
Notice \(9\) is a perfect square integer.
\[(x-4)^2+(y-5)^2~=~3^2\]
Thus,
\[h=4\]\[k=5\]\[r=3\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2+18x+y^2+10y~=~-97\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2+18x+y^2+10y~=~-97\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2+18x+81+y^2+10y+25~=~-97+81+25\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x+9)^2+(y+5)^2~=~9\]
Notice \(9\) is a perfect square integer.
\[(x+9)^2+(y+5)^2~=~3^2\]
Thus,
\[h=-9\]\[k=-5\]\[r=3\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-10x+y^2+12y~=~-45\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-10x+y^2+12y~=~-45\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-10x+25+y^2+12y+36~=~-45+25+36\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-5)^2+(y+6)^2~=~16\]
Notice \(16\) is a perfect square integer.
\[(x-5)^2+(y+6)^2~=~4^2\]
Thus,
\[h=5\]\[k=-6\]\[r=4\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2+8x+y^2+14y~=~16\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2+8x+y^2+14y~=~16\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2+8x+16+y^2+14y+49~=~16+16+49\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x+4)^2+(y+7)^2~=~81\]
Notice \(81\) is a perfect square integer.
\[(x+4)^2+(y+7)^2~=~9^2\]
Thus,
\[h=-4\]\[k=-7\]\[r=9\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-16x+y^2-10y~=~-40\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-16x+y^2-10y~=~-40\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-16x+64+y^2-10y+25~=~-40+64+25\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-8)^2+(y-5)^2~=~49\]
Notice \(49\) is a perfect square integer.
\[(x-8)^2+(y-5)^2~=~7^2\]
Thus,
\[h=8\]\[k=5\]\[r=7\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-4x+y^2+18y~=~-76\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-4x+y^2+18y~=~-76\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-4x+4+y^2+18y+81~=~-76+4+81\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-2)^2+(y+9)^2~=~9\]
Notice \(9\) is a perfect square integer.
\[(x-2)^2+(y+9)^2~=~3^2\]
Thus,
\[h=2\]\[k=-9\]\[r=3\]
Question
Consider the two points:
Point \(A\) is at \((46,86)\)
Point \(B\) is at \((66,36)\)
A third collinear point, \(C\), is 40% of the way from point \(A\) to point \(B\), as shown below. .
Find the coordinates of point \(C\):
(,)
Solution
There are many different ways to approach this problem.
Let’s start by finding the deltas: \(\Delta x\) and \(\Delta y\).
\[\Delta x ~=~ 66-46 ~=~ 20\]\[\Delta y ~=~ 36-86 ~=~ -50\]
The absolute values of these deltas are lengths of legs of a right triangle with vertical and horizontal legs.
We can multiply each absolute delta by the given percentage.
\[0.4 \cdot 20 ~=~ 8\]\[0.4 \cdot 50 ~=~ 20\]
We can draw a new right triangle, starting at \(A\), with these leg lengths. Point \(C\) should be on a vertex of this new triangle.
